Ask The Wizard #326

While this could be solved with a long and tedious Markov chain, I prefer an integral solution. I explain how to use this method in my pages on the Fire Bet and Bonus Craps.

Imagine that instead of significant events being determined by the roll of the die, one at a time, consider them as an instant in time. Assume the time between events has a memory-less property, with an average time between events of one unit of time. In other words, the time between events follows an exponential distribution with a mean of 1. This will not matter for purposes of adjudicating the bet, because events still happen one at a time.

Per the Poisson distribution, the probability that any given side of the die has been rolled zero times in x units of time is exp(-x/6)*(x/6)⁰/0! = exp(-x/6). Poisson also say the probability of any given side being rolled exactly once is exp(-x/6)*(x/6)¹/1! = exp(-x/6) * (x/6). Thus probability any side has been rolled two or more times in x units of time is 1 - exp(-x/6)*(1 + (x/6)). The probability that all six sides have been rolled at least twice is (1 - exp(-x/6)*(1 + (x/6)))⁶. The probability that at least one side has not been rolled at least twice is equal to:

We need to integrate that over all time to find how much time will go by, on average, where the desired goal has not been achieved.

Fortunately, we can use an integral calculator at this point. For the one linked to, put 1- (1 - exp(-x/6)*(1 + x/6))^6 dx = apx. 24.1338692 in the text box following "Calculate the integral of" and under custom, set the bound of integration from 0 to ∞.

The answer is 390968681 / 16200000 = apx. 24.13386919753086

Question 1: 97/6 = apx. 16.166666

Question 2: a⁴/6 + (4/3)ac - a²b + b²/2

For see my solution (PDF)

Let's label the initial die with letters instead of numbers, to avoid confusion. Let's label each possible die state with letters. For example, AAABBC would mean three of one letter, two of another, and one of a third. The initial state would obviously be ABCDEF.

Let E(ABCDEF) be the expected number of rolls from state ABCDEF.

Building on the number of combinations of going from one state to another, the following transition matrix shows how many ways there are for going from each initial state (left column) to each new state. This took a few hours to construct properly, by the way.

I won't go into a long lecture on matrix algebra, except to say let's say matrix B is as follows:

The answer is the determinant of matrix B to that of matrix A:

Determ(A) = 1,461,067,501,120,670,000,000,000,000,000,000,000,000,000,000

Determ(B) = 14,108,055,348,203,100,000,000,000,000,000,000,000,000,000,000

Determ(B) / Determ(A) = apx. 9.65599148388557